@tb I think I figured out why continuous and compactly supported implies uniformly continuous. For an epsilon and a point f(x) in the range there is a delta ball in the domain such that its image is contained in the epsilon ball. The collection of all these delta balls covers the domain and that's compact so has a finite subcover. Then choosing the smallest delta of all these gives the delta that makes it uniformly continuous.
@JM Her Majesty will like it! Fern by a British mathematician makes it twice as appealing!
@Matt Yes, that's the basic idea. However, there's some fleshing out to do: why exactly is it that any two points that aren't more than delta apart are contained in the same ball? (that's actually not true, so you should phrase this a bit more carefully)
None of my torus experiments the past few days came out the way I wanted 'em (but maybe I'll use them in the future), so I got lazy and used one of my old Christmas avatar standbys...
That's true. I am glad they came out with the Home Edition. I used to get a similarly priced educational edition from Academic Superstore, but getting it directly from Wolfram is good (in case I am without university discounts at some point).
@tb Oh, I see. The only thing that I missed was that my delta balls weren't necessarily subsets of the delta_i balls in the finite subcover and therefore the f values not necessarily contained in the epsilon balls. Nice.
@Matt If you are talking about the Lebesgue number problem, I think I remember proving something with the maximum of the set of distance functions to the complements of each open set in the cover. Note that the maximum of a finite collection of continuous functions is continuous. Find the minimum of this maximum and note that it is >0.
@Matt Choose $\epsilon>0$. Since the function is continuous, at each point $x$, there is a ball $B_x$ of radius $r_x$ centered at $x$ so that $|f(y)-f(x)|<\epsilon$ for all $y\in B_x$.
@robjohn I did that. Although now that I come to think of it I'm not sure why I need the finite subcover. I can just pick the smallest delta in the original cover.
@robjohn I thought I had figured that out: Pick the smallest delta in the finite subcover and then the delta balls are in the delta_i balls of the finite subcover so all the f values are epsilon close together.
@Matt: Regarding your dating advice from the other day, I woke up this morning and remembered that between my first and second year as an undergrad I hooked up with this girl from my classes (though she was in comp. sci.) and later I found out that she was dating someone and were in an open relationship at the time.
@Matt Well, I was hoping to hook up with her again but by the time I found out they already closed the relationship and outed it (it was a bit secret at first...)
@Matt another way to work it is to cover the compact set with balls of half the radius needed to insure that $f$ in the ball differs less than $\epsilon/2$ from $f$ at the center.
@Matt then let $\delta$ be the smallest of the radii of the finite subcover chosen.
@Matt pick any $|x-y|<\delta$ and then both are within $2\delta$ of any ball containing $x$ thus, $|f(x)-f(y)|<|f(x)-f(\text{center})|+|f(y)-f(\text{center})|<\epsilon$
@Matt choosing the balls to be half the radius required for $\epsilon/2$ is important to this method.
